Open Query-Assignment 17

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course Phy 232

7/15 11

Question: `q**** query univ phy problem 33.44 11th edition 33.38 (34.28 10th edition) 3 mm plate, n = 1.5, in 3 cm separation between 450 nm source and screen. How many wavelengths are there between the source and the screen?

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Your solution:

The wavelength in glass is 450nm / 1.5 = 300nm.

I am unsure how many wavelengths are between the source and screen.

confidence rating #$&*:

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Given Solution:

`a** The separation consists of 1.5 cm = 1.5 * 10^7 nm of air, index of refraction very close to 1, and 1.5 mm = 1.5 * 10^6 nm of glass, index of refraction 1.5.

The wavelength in the glass is 450 nm / 1.5 = 300 nm, approx..

So there are 1.5 * 10^7 nm / (450 nm/wavelength) = 3.3 * 10^4 wavelengths in the air and 1.5 * 10^6 nm / (300 nm/wavelength) = 5.0 * 10^3 wavelengths in the glass. **

STUDENT QUESTION

After reading the solution, I am unsure about why the path difference is equal to twice the plate thickness.

INSTRUCTOR RESPONSE

One ray is reflected from the 'top' surface of the plate. The other passes through the plate to the other surface, is reflected there. It has already passed through the thickness of the plate. It has to pass back through the plate, to the other surface, before it 'rejoins' the original ray. It has therefore traveled an extra distance equal to double that of the plate.

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Self-critique (if necessary):

Why is the separation n * 10^7? I do not understand this index of refraction.

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Self-critique Rating: 2

@& I don't see anything in the given solution about n * 10^7. (That doesn't mean it's there; I could be missing it). In any case I'm not sure what your question is.

However, I'll risk tellilng you what you might already know:

nm is the abbreviation for nanometer. A nanometer is 10^-9 meter, and therefore 10^-7 centimeters.

So there are 10^7 nm in a cm.

The light travels throug 3 mm = .3 cm of glass, while its wavelength is 300 nm, and 3 cm of air, while its wavelength is 450 nm.*@

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&#This looks good. See my notes. Let me know if you have any questions. &#